A farmer has 200 acres of land suitable for cultivating crops A, B and C. The cost per acre of cultivating crop A, crop B and crop C is $40, $60 and $80, respectively. The farmer has $12,600 available for land cultivation. Each acre of crop A requires 20 labor-hours, each acre of crop B requires 25 labor-hours and each acre of crop C requires 40 labor-hours. The farmer has a maximum of 5950 labor-hours available. If he wishes to use all his cultivatable land, the entire budget and all the labor available, how many acres of each crop should he plant? [Please can you show me how to Use the Gauss Jordan Elimination method to solve]

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To use the Gauss Jordan Elimination method to solve this problem, we will first set up a system of equations based on the information given.

Let's assume the farmer plants x acres of crop A, y acres of crop B, and z acres of crop C.

The first constraint is the total number of acres:
x + y + z = 200

The second constraint is the total budget:
40x + 60y + 80z = 12,600

The third constraint is the total labor hours:
20x + 25y + 40z = 5950

Now, we can solve this system of equations using the Gauss Jordan Elimination method.

Step 1: Set up the augmented matrix:
| 1 1 1 | 200 |
| 40 60 80 | 12,600 |
| 20 25 40 | 5950 |

Step 2: Perform row operations to reduce the matrix into row-echelon form:

- Divide row 2 by 40:
| 1 1 1 | 200 |
| 1.5 2.25 3 | 315 |
| 20 25 40 | 5950 |

- Multiply row 3 by (-1/20) and add it to row 1:
| 0 1 3/4 | 130 |
| 1.5 2.25 3 | 315 |
| 0 1/4 1 | 180 |

- Multiply row 3 by (-2/3) and add it to row 2:
| 0 1 3/4 | 130 |
| 1.5 0 -1/2 | 185 |
| 0 1/4 1 | 180 |

- Multiply row 2 by (4/3) and subtract row 1:
| 0 1 0 | 10 |
| 0 0 -1 | 320 |
| 0 1/4 1 | 180 |

- Multiply row 3 by (1/4) and subtract row 2:
| 0 1 0 | 10 |
| 0 0 -1 | 320 |
| 0 0 1 | 20 |

- Multiply row 2 by (-1) and add it to row 3:
| 0 1 0 | 10 |
| 0 0 -1 | 320 |
| 0 0 0 | -300 |

Step 3: Continue row operations to reduce the matrix into reduced row-echelon form:

- Multiply row 3 by (-1/300):
| 0 1 0 | 10 |
| 0 0 1 | -320 |
| 0 0 0 | 1 |

Since the last row indicates that 0 = 1, the system is inconsistent, meaning there is no solution that satisfies all the constraints.

Therefore, it is not possible for the farmer to use all his cultivatable land, the entire budget and all the labor available.

To solve this problem using Gauss Jordan Elimination, we can set up a system of linear equations based on the given information.

Let's denote the number of acres of crop A, B, and C as x, y, and z respectively.

1. The first equation represents the total cost of cultivation:
40x + 60y + 80z = 12600

2. The second equation represents the total labor-hours required:
20x + 25y + 40z = 5950

3. The third equation represents the total land available:
x + y + z = 200

To solve these equations using Gauss Jordan Elimination, we can set up an augmented matrix:

| 40 60 80 | 12600 |
| 20 25 40 | 5950 |
| 1 1 1 | 200 |

Now, let's perform Gauss Jordan Elimination step-by-step to solve the system:

Step 1: Divide the first row by 40 to make the leading coefficient of the first row 1:
| 1 1.5 2 | 315 |
| 20 25 40 | 5950 |
| 1 1 1 | 200 |

Step 2: Replace the second row by (second row - 20 * first row):
| 1 1.5 2 | 315 |
| 0 5 20 | 250 |
| 1 1 1 | 200 |

Step 3: Replace the third row by (third row - first row):
| 1 1.5 2 | 315 |
| 0 5 20 | 250 |
| 0 -0.5 -1 | -115 |

Step 4: Divide the second row by 5 to make the leading coefficient of the second row 1:
| 1 1.5 2 | 315 |
| 0 1 4 | 50 |
| 0 -0.5 -1 | -115 |

Step 5: Replace the first row by (first row - 1.5 * second row) and the third row by (third row + 0.5 * second row):
| 1 0 -1 | 240 |
| 0 1 4 | 50 |
| 0 0 -1 | -65 |

Step 6: Multiply the third row by -1 to get the leading coefficient of the third row 1:
| 1 0 -1 | 240 |
| 0 1 4 | 50 |
| 0 0 1 | 65 |

Step 7: Replace the first row by (first row + second row):
| 1 1 3 | 290 |
| 0 1 4 | 50 |
| 0 0 1 | 65 |

Step 8: Replace the first row by (first row - 3 * third row) and the second row by (second row - 4 * third row):
| 1 1 0 | 95 |
| 0 1 0 | 10 |
| 0 0 1 | 65 |

Step 9: Replace the first row by (first row - second row):
| 1 0 0 | 85 |
| 0 1 0 | 10 |
| 0 0 1 | 65 |

The final matrix represents the solution to the system of equations:
x = 85
y = 10
z = 65

Therefore, the farmer should plant 85 acres of crop A, 10 acres of crop B, and 65 acres of crop C to utilize all his cultivatable land, the entire budget, and all the available labor.